2. In areas of the world where outdoor temperatures are low enough to drop the interior building temperature below setpoint.

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1 In addition to a building s cooling requirements, there is typically a need to provide some type of heating capability to maintain setpoint temperatures. This could be: 1. Upon building warmup after a period where the systems do not operate i.e. on Monday mornings after the systems have allowed the building to operate below setpoint to save energy. 2. In areas of the world where outdoor temperatures are low enough to drop the interior building temperature below setpoint. 3. In humid regions where the air handling systems are required to produce extremely air temperatures for moisture control. Building Heat Load Calculation Later in the course, we will be focused on building load calculations. For now, realize the several factors impact the overall heating and cooling loads on a building. These are: Infiltration air that comes into a building around doors and windows, and through gaps in the construction. Outside air/ventilation air outside ambient air needed to meet code requirements and to maintain building pressurization. Fenestration/glass glass in doors, windows, skylights, etc. can have a large impact on a buildings heating and cooling requirements. Duct heat gain/heat loss as ductwork passes through unconditioned spaces (i.e. attics, plenums, equipment rooms), it tends to pick up or lose heat. Appliance heat gain mechanical and electrical equipment within a building impacts the heating and cooling loads by contributing to the overall heat load of the building. Occupant heat gain people working in or visiting a building contribute to the heat load. Even people sitting and not moving have an impact. Design temperatures indoor and outdoor design temperatures for the building s location significantly impact the final heating and EngineeringDesignResources.com 1

2 cooling heat load calculation and the equipment installed to condition the air. Impact of Geographic Location on HVAC Systems Consider the following 2 scenarios. Building 1 Located in a moderate climate zone where temperatures rarely drop below 45 F. Given this fact, the cooling design will dictate the air handling unit sizes. If the cooling load for this building is 2,000,000 BTUh with a design temperature difference of 20 F, the required airflow would be: CFM = (cooling load) (1.08 x ΔT) = (2,000,000) (1.08 x 20) CFM = 92,593 cfm If the heating load for this building is only 1,000,000 BTUh, what is the heating airflow required? (Assume a mixed air temperature of 67 F and a required heating coil leaving air temperature of 95 F) CFM = (heating load) (1.08 x ΔT) = (1,000,000) (1.08 x [95 67]) CFM = 33,069 cfm That s a difference of 59,524 cfm. This difference in airflow requirement for heating and cooling would require a variable flow system or some other configuration to adequately heat and cool the building. Building 2 Identical to building 1 but is located in a cold climate zone where temperatures consistently drop below 0 F. If the cooling load for this building is 1,000,000 BTUh with a design temperature difference of 20 F, the required airflow would be: CFM = (cooling load) (1.08 x ΔT) = (1,000,000) (1.08 x 20) CFM = 46,296 cfm EngineeringDesignResources.com 2

3 The heating load for this building is 2,000,000 BTUh, what is the heating airflow required? (Assume a mixed air temperature of 58 F and a required heating coil leaving air temperature of 95 F) CFM = (heating load) (1.08 x ΔT) = (2,000,000) (1.08 x [95 58]) CFM = 50,050 cfm That s a difference of only 3,754 cfm. This system could be manipulated to a constant volume system if necessary. But notice, although very close, the heating cfm dictates the air handling unit size. Download: Psychrometric Chart Download: The Psychrometric Chart and the Air Conditioning Process Familiarize yourself with: The saturation line Lines of constant relative humidity Lines of constant wet bulb temperature Lines of constant specific volume Lines of constant enthalpy Download: The Psychrometric Chart: Theory and Application This presentation is in metric but we will be using the concepts discussed in the presentation. Be familiar with all the parts of the Psychrometric Chart. EngineeringDesignResources.com 3

4 Practice Problems Problem 1 5,000 cfm of outside air at 98 Fdb/85 Fwb is mixed with 20,000 cfm of return air at 72 F/50%RH. What is the mixed air temperature and relative humidity? Solution a. Find the specific volumes: At 98 Fdb/85 Fwb, SpV = ft 3 /lb of dry air At 72 F/50%RH, SpV = ft 3 /lb of dry air b. Convert CFM to pounds of air: 5,000 cfm ft 3 /lb = lbs/min 20,000 cfm ft 3 /lb = lbs/min c. Find the total weight of the mixed air: lbs/min lbs/min = lbs/min d. Calculate the dry bulb temperature: Tdb O/A = ( ) x 98 Fdb = Fdb Tdb R/A = ( ) x 72 Fdb = Fdb Tdb mixed air = = Fdb e. Draw a line between the 2 points (98 Fdb/85 Fwb & 72 F/50%RH.) Locate the intersection of this line and Fdb. Determine the relative humidity. RH = 58% f. Determine the wet bulb temperature. Twb = 66.2 Fwb EngineeringDesignResources.com 4

5 Problem 2 Determine the specific humidity, the enthalpy, the dew point temperature, and the specific volume of the mixed air in problem 1. Solution a. Specific humidity: Draw a horizontal line to the right through the point Fdb & 66.2 Fwb. Determine where this horizontal line intersects the Humidity Ratio/Specific Humidity scale. Specific humidity (or Humidity ratio) = 78.2 grains/lbdry air b. Enthalpy: Draw a line through the point Fdb & 66.2 Fwb that intersects the 2 enthalpy scales (1 along the left side of the psychrometric chart and 1 along the bottom of the psychrometric chart) at the same number. (Refer to sketch.) Enthalpy = 30.7 Btu/lbdry air c. Dewpoint temperature: Extend the horizontal line (drawn to determine specific humidity) to the right to intersect the Dewpoint Temperature scale. Dewpoint temperature = 60.5 F d. Specific volume: Determine Specific Volume from location of point Fdb & 66.2 Fwb on the psychrometric chart. SpV = ft 3 /lbdry air EngineeringDesignResources.com 5

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7 Problem 3 3,000 cfm of outside air at 65 Fdb/40%RH is mixed with 3,000 cfm of return air at 75 F/50%RH. What is the mixed air temperature and relative humidity? Solution a. Find the specific volumes: At 65 Fdb/40%RH, SpV = ft 3 /lb of dry air At 75 F/50%RH, SpV = ft 3 /lb of dry air b. Convert CFM to pounds of air: 3,000 cfm ft 3 /lb = lbs/min 3,000 cfm ft 3 /lb = lbs/min c. Find the total weight of the mixed air: lbs/min lbs/min = lbs/min d. Calculate the dry bulb temperature: Tdb O/A = ( ) x 65 Fdb = Fdb Tdb R/A = ( ) x 75 Fdb = Fdb Tdb mixed air = = Fdb e. Draw a line between the 2 points (65 Fdb/40%RH & 75 F/50%RH.) Locate the intersection of this line and Fdb. Determine the relative humidity. RH = 46% EngineeringDesignResources.com 7

8 Problem 4 If the air streams from problem 3 are serving a space that requires a constant temperature of 72 Fdb and a relative humidity of 60%RH, how much heat (BTUh) must be added to the supply air? How much moisture (water, gal/hr)? Solution a. Find the enthalpies: For Fdb/46%RH, h1 = 24.6 Btu/lbdry air For 72 Fdb/60%RH, h2 = 28.2 Btu/lbdry air b. Total heat required: Qh = h2 h1 = 28.2 Btu/lbdry air 24.6 Btu/lbdry air = 3.6 Btu/lbdry air Qh = (3.6 Btu/lbdry air)x( lbs/min)x(60 min/hr) = 95,982 BTUh c. Find the Specific Humidities: For Fdb/46%RH, specific humidity = 50 grains/lbdry air For 72 Fdb/60%RH, specific humidity = 71 grains/lbdry air d. Total moisture required = 71 grains/lbdry air - 50 grains/lbdry air (21 grains/lbdry air)x( lbs/min)x(60 min/hr) = 559,894 grains/hr e. Convert grains/hr to lbs/hr: 559,894 grains/hr 7000 grains/lb = 80 lbs/hr f. Convert lbs/hr to gallons/hr: 80 lbs/hr 8.33 lbs/gal = 9.6 gal/hr EngineeringDesignResources.com 8